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a cylindrical water tank with its circular base parallel to the ground is being filled at the rate of 4 cubic feet per minute. the radius of the tank is 2 feet. how fast is the level of the water in the tank rising when the tank is half full?

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Answer:

To solve this problem, we need to use the formula for the volume of a cylinder: V = πr^2h, where V is the volume, r is the radius, and h is the height.

Step 1: We know that the radius of the tank is 2 feet and that it is being filled at a rate of 4 cubic feet per minute. We also know that the tank is half full, so we can assume that the height of the water in the tank is half the height of the tank itself.

Step 2: We can use the volume formula to find the volume of the tank when it is half full.

V = πr^2h

V = π(2^2)(h/2)

V = 4πh/2

Step 3: We can divide the rate of fill by the volume of the tank when it is half full to find the rate of change of the height of the water.

dh/dt = 4/(4π/2)

dh/dt = 2/π ft/min

Final Answer: The level of the water in the tank is rising at a rate of 2/π feet per minute when the tank is half full.

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